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Maximum Sum Decreasing Subsequence
  • Difficulty Level : Easy
  • Last Updated : 05 May, 2021

Given an array of N positive integers. The task is to find the sum of the maximum sum decreasing subsequence(MSDS) of the given array such that the integers in the subsequence are sorted in decreasing order. 
Examples

Input: arr[] = {5, 4, 100, 3, 2, 101, 1} 
Output: 106 
100 + 3 + 2 + 1 = 106
Input: arr[] = {10, 5, 4, 3} 
Output: 22 
10 + 5 + 4 + 3 = 22 
 

This problem is a variation of the Longest Decreasing Subsequence problem. The Optimal Substructure for the above problem will be: 
Let arr[0..n-1] be the input array and MSDS[i] be the maximum sum of the MSDS ending at index i such that arr[i] is the last element of the MSDS. 
Then, MSDS[i] can be written as: 

MSDS[i] = a[i] + max( MSDS[j] ) where i > j > 0 and arr[j] > arr[i] or, 
MSDS[i] = a[i], if no such j exists.

To find the MSDS for a given array, we need to return max(MSDS[i]) where n > i > 0. 
Below is the implementation of the above approach: 
 

C++




// CPP code to return the maximum sum
// of decreasing subsequence in arr[]
#include <bits/stdc++.h>
using namespace std;
 
// function to return the maximum
// sum of decreasing subsequence
// in arr[]
int maxSumDS(int arr[], int n)
{
    int i, j, max = 0;
    int MSDS[n];
 
    // Initialize msds values
    // for all indexes
    for (i = 0; i < n; i++)
        MSDS[i] = arr[i];
 
    // Compute maximum sum values
    // in bottom up manner
    for (i = 1; i < n; i++)
        for (j = 0; j < i; j++)
            if (arr[i] < arr[j] && MSDS[i] < MSDS[j] + arr[i])
                MSDS[i] = MSDS[j] + arr[i];
 
    // Pick maximum of all msds values
    for (i = 0; i < n; i++)
        if (max < MSDS[i])
            max = MSDS[i];
 
    return max;
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 4, 100, 3, 2, 101, 1 };
     
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Sum of maximum sum decreasing subsequence is: "
         << maxSumDS(arr, n);
    return 0;
}

Java




// Java code to return the maximum sum
// of decreasing subsequence in arr[]
import java.io.*;
import java.lang.*;
 
class GfG {
     
    // function to return the maximum
    // sum of decreasing subsequence
    // in arr[]
    public static int maxSumDS(int arr[], int n)
    {
        int i, j, max = 0;
        int[] MSDS = new int[n];
     
        // Initialize msds values
        // for all indexes
        for (i = 0; i < n; i++)
            MSDS[i] = arr[i];
     
        // Compute maximum sum values
        // in bottom up manner
        for (i = 1; i < n; i++)
            for (j = 0; j < i; j++)
                if (arr[i] < arr[j] &&
                    MSDS[i] < MSDS[j] + arr[i])
                    MSDS[i] = MSDS[j] + arr[i];
     
        // Pick maximum of all msds values
        for (i = 0; i < n; i++)
            if (max < MSDS[i])
                max = MSDS[i];
     
        return max;
    }
     
    // Driver Code
    public static void main(String argc[])
    {
        int arr[] = { 5, 4, 100, 3, 2, 101, 1 };
         
        int n = 7;
     
        System.out.println("Sum of maximum sum"
               + " decreasing subsequence is: "
                           + maxSumDS(arr, n));
    }
}
 
// This code os contributed by Sagar Shukla.

Python3




# Python3 code to return the maximum sum
# of decreasing subsequence in arr[]
 
# Function to return the maximum
# sum of decreasing subsequence
# in arr[]
def maxSumDS(arr, n):
     
    i, j, max = (0, 0, 0)
     
    MSDS=[0 for i in range(n)]
  
    # Initialize msds values
    # for all indexes
    for i in range(n):
        MSDS[i] = arr[i]
  
    # Compute maximum sum values
    # in bottom up manner
    for i in range(1, n):
        for j in range(i):
            if (arr[i] < arr[j] and
                MSDS[i] < MSDS[j] + arr[i]):
                MSDS[i] = MSDS[j] + arr[i]
  
    # Pick maximum of all msds values
    for i in range(n):
        if (max < MSDS[i]):
            max = MSDS[i]
 
    return max
     
if __name__ == "__main__":
     
    arr=[5, 4, 100, 3,
         2, 101, 1]
    n=len(arr)
    print("Sum of maximum sum decreasing subsequence is: ",
           maxSumDS(arr, n))
 
# This code is contributed by Rutvik_56

C#




// C# code to return the
// maximum sum of decreasing
// subsequence in arr[]
using System;
 
class GFG
{
     
    // function to return the
    // maximum sum of decreasing
    // subsequence in arr[]
    public static int maxSumDS(int []arr,
                               int n)
    {
        int i, j, max = 0;
        int[] MSDS = new int[n];
     
        // Initialize msds values
        // for all indexes
        for (i = 0; i < n; i++)
            MSDS[i] = arr[i];
     
        // Compute maximum sum values
        // in bottom up manner
        for (i = 1; i < n; i++)
            for (j = 0; j < i; j++)
                if (arr[i] < arr[j] &&
                    MSDS[i] < MSDS[j] + arr[i])
                    MSDS[i] = MSDS[j] + arr[i];
     
        // Pick maximum of
        // all msds values
        for (i = 0; i < n; i++)
            if (max < MSDS[i])
                max = MSDS[i];
     
        return max;
    }
     
    // Driver Code
    static public void Main ()
    {
        int []arr = {5, 4, 100,
                     3, 2, 101, 1};
        int n = 7;
        Console.WriteLine("Sum of maximum sum" +
                " decreasing subsequence is: " +
                              maxSumDS(arr, n));
    }
}
 
// This code is contributed by m_kit

PHP




<?php
// PHP code to return the maximum sum
// of decreasing subsequence in arr[]
 
// function to return the maximum
// sum of decreasing subsequence
// in arr[]
function maxSumDS($arr, $n)
{
    $i; $j; $max = 0;
    $MSDS = array();
 
    // Initialize msds values
    // for all indexes
    for ($i = 0; $i < $n; $i++)
        $MSDS[$i] = $arr[$i];
 
    // Compute maximum sum values
    // in bottom up manner
    for ($i = 1; $i < $n; $i++)
        for ($j = 0; $j < $i; $j++)
            if ($arr[$i] < $arr[$j] &&
                   $MSDS[$i] < $MSDS[$j] + $arr[$i])
                $MSDS[$i] = $MSDS[$j] + $arr[$i];
 
    // Pick maximum of
    // all msds values
    for ($i = 0; $i < $n; $i++)
        if ($max < $MSDS[$i])
            $max = $MSDS[$i];
 
    return $max;
}
 
// Driver Code
$arr = array (5, 4, 100,
              3, 2, 101, 1 );
 
$n = sizeof($arr);
 
echo "Sum of maximum sum decreasing " .
                    "subsequence is: ",
                    maxSumDS($arr, $n);
 
// This code is contributed by ajit
?>

Javascript




<script>
    // Javascript code to return the
    // maximum sum of decreasing
    // subsequence in arr[]
     
    // function to return the
    // maximum sum of decreasing
    // subsequence in arr[]
    function maxSumDS(arr, n)
    {
        let i, j, max = 0;
        let MSDS = new Array(n);
      
        // Initialize msds values
        // for all indexes
        for (i = 0; i < n; i++)
            MSDS[i] = arr[i];
      
        // Compute maximum sum values
        // in bottom up manner
        for (i = 1; i < n; i++)
            for (j = 0; j < i; j++)
                if (arr[i] < arr[j] &&
                    MSDS[i] < MSDS[j] + arr[i])
                    MSDS[i] = MSDS[j] + arr[i];
      
        // Pick maximum of
        // all msds values
        for (i = 0; i < n; i++)
            if (max < MSDS[i])
                max = MSDS[i];
      
        return max;
    }
     
    let arr = [5, 4, 100, 3, 2, 101, 1];
    let n = 7;
    document.write("Sum of maximum sum" +
                      " decreasing subsequence is: " +
                      maxSumDS(arr, n));
 
// This code is contributed by suresh07.
</script>

Output
 

Sum of maximum sum decreasing subsequence is: 106

Time complexity: O(N2
Auxiliary Space: O(N)




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